Chemistry · Redox Equilibrium · NEET
An oxidising agent is a species that takes electrons from something else, so it gets reduced itself. In K2Cr2O7 the chromium is in a high +6 state. It really wants to drop to the more stable +3 state, and to do that it must pull electrons from another substance. When it pulls those electrons, the other substance is oxidised. So K2Cr2O7 oxidises others and is reduced itself — that is exactly what makes it an oxidising agent.
In K2Cr2O7: K is +1 (two of them = +2), each O is -2 (seven of them = -14). The molecule is neutral, so the two Cr atoms together must be +12, meaning each Cr is +6. After the reaction Cr becomes +3. So each chromium falls from +6 to +3, a drop of 3 per atom, and 6 in total for the two Cr atoms.
The orange colour comes from the dichromate ion Cr2O7^2- (Cr is +6). When it acts as an oxidising agent, Cr is reduced to Cr^3+ ions (like in Cr2(SO4)3). The Cr^3+ ion is green. So as +6 chromium turns into +3 chromium, the solution changes from orange to green. This colour change is a common NEET clue that dichromate has done its oxidising job.
Six. The half-reaction is: Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O. There are two Cr atoms, each going from +6 to +3 (gaining 3 electrons each), so 2 x 3 = 6 electrons total. This is why you always multiply the other half-reaction to supply exactly 6 electrons when balancing.
The reduction needs H+ ions. Look at the half-reaction: the seven oxygen atoms in Cr2O7^2- leave as water, and that requires 14 H+ ions to form 7 H2O. Without acid (usually dilute H2SO4) there are not enough H+ ions, so the reaction cannot go forward. That is why NEET always says 'acidified K2Cr2O7'.
KMnO4 is generally the stronger oxidising agent (its standard potential is higher, about +1.51 V vs about +1.33 V for dichromate). But K2Cr2O7 has one big practical advantage: it can be used as a primary standard because it is stable, pure and does not react with dilute HCl the way KMnO4 does. So KMnO4 = stronger, K2Cr2O7 = more reliable for titrations.
Which one of the following statements is correct when SO2 is passed through acidified K2Cr2O7 solution?
On balancing the redox reaction, the coefficients a, b and c are, respectively: a Cr2O7^2- + b SO3^2- + c H+ -> 2Cr^3+ + b SO4^2- + (c/2) H2O
The oxidation state of Cr in CrO5 is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
In acidic medium: Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O. Two Cr atoms each fall from +6 to +3, gaining 6 electrons in total.
Because Cr^6+ (orange Cr2O7^2-) is reduced to Cr^3+, which is green. The colour change orange -> green is the visible sign that dichromate has acted as an oxidising agent.
Yes. Acidified dichromate oxidises Fe2+ to Fe3+ and is reduced to Cr3+. Six Fe2+ ions supply the 6 electrons that one Cr2O7^2- needs. This is a classic redox titration used to estimate iron.
K2Cr2O7 is stable, can be obtained pure and dry, does not decompose in light, and does not react with dilute HCl. KMnO4 is less stable and reacts with HCl, so it cannot be a primary standard.
